Question

tan of Cos inverse 4 by 5 + tan inverse 2 by 3

Answer 1

Here is solution.......

Answer 2

$i )Let \: x = cos^{-1} \big(frac{4}{5} \big)$

$\implies cosx = \frac{4}{5} \: --(1)$

$ii ) sin^{2} x = 1 - cos^{2} x \\= 1 - \big(\frac{4}{5} \big)^{2} \\= 1 - \frac{16}{25} \\= \frac{25-16}{25} \\= \frac{9}{25}\\= \big(\frac{3}{5}\big)^{2}$

$Now , sin x = \frac{3}{5} \:--(2)$

$iii) tan x = \frac{sin x}{cos x } \\= \frac{\frac{3}{5}}{\frac{4}{5}} \\= \frac{3}{5} \: --(3)$

$iv) \red{ Value \: of \: tan \Big( cos^{-1} \big(\frac{4}{5}\big) + tan^{-1}\big(\frac{2}{3}\big) \Big) }$

$= tan \Big( tan^{-1} \frac{3}{4} + tan^{-1} \frac{2}{3} \Big)$

$= tan\Big(tan^{-1} \frac{\frac{3}{4} + \frac{2}{3}}{ 1 - \frac{3}{4} \times \frac{2}{3}} \Big)$

$= \frac{ \frac{3}{4} + \frac{2}{3}}{ 1 - \frac{3}{4} \times \frac{2}{3} } \\= \frac{\frac{9+8}{12}}{\frac{12-6}{12}}$

$= \frac{17}{6}$

Therefore.,

$\red{ Value \: of \: tan \Big( cos^{-1} \big(\frac{4}{5}\big) + tan^{-1}\big(\frac{2}{3}\big) \Big) }$

$\green {= \frac{17}{6} }$

•••♪